and you want to solve them using the elimination method, you
must change their coefficients first, so they are opposites of each other (that
is even what happens in subtracting equations). What did you do when you
subtracted? You multiplied everything in one equation by -1. In
these equations, we must simply multiply by something else.

Sometimes the equations may have one coefficient twice that of
the same variable in the other equation (e.g. x + y = 9, 2x - y = 6), and it is
easy to multiply the first equation by -2 (remember, get them to be opposites).
If you feel like trying that one, the solution is (5, 4). Systems like the
first one, however, require both equations to be multiplied. To do this,
first find the least common multiple of the coefficients of whichever variable
you want to eliminate. You use the least common multiple because you want
to work with as small numbers as possible. Your choice of which variable
to eliminate may also be influenced by with least common multiple is smaller
(and when I say smaller, I mean a smaller absolute value).

The lower least common multiple is 12, and will eliminate x.
You will multiply the first equation by either 4 or -4, and the second by -3 or
3, respectively.

4(3x + 5y = 12)
12x + 20y = 48

-3(4x - 3y = -13)
-12x + 9y = 39

12x + 20y = 48-12x + 9y = 39 29y = 87

y = 3
3x + 5(3) = 12
3x + 15 = 12
3x = -3
x = -1

The solution to these equations is (-1, 3).

Try another:

4x + 10y = 50
2x + 5y = 25

You can just multiply the second equation by -2 to get one that
can be added to the first.

2(2x + 5y = 25)
-4x - 10y = -50

You can probably, without adding the equations, realize the
result will be 0 = 0. As you know, if you get a true statement as a
result, the equations are dependent, and all their solutions are common.

Similarly, if the second equation had been 2x + 5y = 20, when
multiplied by -2 it would have been -4x - 10y = -40. When added to the
first equation, it would give 0 = 10, a false statement. That false
statement shows that the equations are inconsistent, with no common solutions.

Now that you know two ways to solve systems of equations
(without the effort of making graphs), you must decide when to use them.
It is up to you, but here are a few helpful hints:

If one variable has no coefficient on it, it is easy to use
substitution (substitute for it), as in x + 4y = 12, -2x + 16y = 0. You
can easily solve for x in the first equation, so it lends itself to
substitution. By the way, the solution to this is (8, 1).

If you have an equation where the coefficients of one or both
variables are the same or opposites, it is easy to use the elimination method,
rather than use fractions in substitution.